
What is the adjoint of \[A = \left[ {\begin{array}{*{20}{c}}3&{ - 3}&4\\2&{ - 3}&4\\0&{ - 1}&1\end{array}} \right]\]?
A. \[\left[ {\begin{array}{*{20}{c}}4&8&3\\2&1&6\\0&2&1\end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}1&{ - 1}&0\\{ - 2}&3&{ - 4}\\{ - 2}&3&{ - 3}\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}{11}&9&3\\1&2&8\\6&9&1\end{array}} \right]\]
D. \[\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\{ - 1}&3&3\\{ - 2}&3&{ - 3}\end{array}} \right]\]
Answer
163.2k+ views
Hint: We will find the cofactors of the given matrix. Then we will find the transpose matrix of the cofactor matrix to find the adjoint matrix.
Formula used:
The adjoint of the matrix \[\left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right]\] is \[\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{21}}}&{{A_{31}}}\\{{A_{12}}}&{{A_{22}}}&{{A_{32}}}\\{{A_{13}}}&{{A_{23}}}&{{A_{33}}}\end{array}} \right]\]where \[{A_{ij}}\] are the cofactors.
Complete step by step solution:
Given matrix is \[A = \left[ {\begin{array}{*{20}{c}}3&{ - 3}&4\\2&{ - 3}&4\\0&{ - 1}&1\end{array}} \right]\].
The cofactors of A are
\[{A_{11}} = \left| {\begin{array}{*{20}{c}}{ - 3}&4\\{ - 1}&1\end{array}} \right|\]
\[ = - 3 \cdot 1 - 4\left( { - 1} \right)\]
\[ = 1\]
\[{A_{12}} = - \left| {\begin{array}{*{20}{c}}2&4\\0&1\end{array}} \right|\]
\[ = - \left( {2 \cdot 1 - 4 \cdot 0} \right)\]
\[ = - 2\]
\[{A_{13}} = \left| {\begin{array}{*{20}{c}}2&3\\0&{ - 1}\end{array}} \right|\]
\[ = 2 \cdot \left( { - 1} \right) - 3 \cdot 0\]
\[ = - 2\]
\[{A_{21}} = - \left| {\begin{array}{*{20}{c}}{ - 3}&4\\{ - 1}&1\end{array}} \right|\]
\[ = - \left[ { - 3 \cdot 1 - 4 \cdot \left( { - 1} \right)} \right]\]
\[ = - 1\]
\[{A_{22}} = \left| {\begin{array}{*{20}{c}}3&4\\0&1\end{array}} \right|\]
\[ = \left[ {3 \cdot 1 - 4 \cdot 0} \right]\]
\[ = 3\]
\[{A_{23}} = - \left| {\begin{array}{*{20}{c}}3&{ - 3}\\0&{ - 1}\end{array}} \right|\]
\[ = - \left[ {3 \cdot \left( { - 1} \right) - \left( { - 3} \right) \cdot 0} \right]\]
\[ = 3\]
\[{A_{31}} = \left| {\begin{array}{*{20}{c}}{ - 3}&4\\{ - 3}&4\end{array}} \right|\]
\[ = \left[ { - 3 \cdot 4 - \left( { - 3} \right) \cdot 4} \right]\]
\[ = 0\]
\[{A_{32}} = - \left| {\begin{array}{*{20}{c}}3&4\\2&4\end{array}} \right|\]
\[ = - \left[ {3 \cdot 4 - 2 \cdot 4} \right]\]
\[ = - 4\]
\[{A_{33}} = \left| {\begin{array}{*{20}{c}}3&{ - 3}\\2&{ - 3}\end{array}} \right|\]
\[ = \left[ {3 \cdot \left( { - 3} \right) - 2 \cdot \left( { - 3} \right)} \right]\]
\[ = - 3\]
The Adjoint of the matrix is
\[Adj\,A = \left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{21}}}&{{A_{31}}}\\{{A_{12}}}&{{A_{22}}}&{{A_{32}}}\\{{A_{13}}}&{{A_{23}}}&{{A_{33}}}\end{array}} \right]\]
Substitute the value of cofactors
\[ = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&0\\{ - 2}&3&{ - 4}\\{ - 2}&3&{ - 3}\end{array}} \right]\]
Hence option B is the correct option.
Additional information:
We can find the adjoint of a matrix if the matrix is a square matrix. This means the number of rows is equal to the number of columns. The adjoint of a matrix is used to the determine inverse of that matrix. In other words, we can find the inverse of a matrix if the matrix is a square matrix.
Note: Students often do a common mistake to find the adjoint of the matrix. They forgot to transpose the cofactor matrix. To calculate the adjoint we have to find the transpose matrix of the cofactor matrix.
Formula used:
The adjoint of the matrix \[\left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right]\] is \[\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{21}}}&{{A_{31}}}\\{{A_{12}}}&{{A_{22}}}&{{A_{32}}}\\{{A_{13}}}&{{A_{23}}}&{{A_{33}}}\end{array}} \right]\]where \[{A_{ij}}\] are the cofactors.
Complete step by step solution:
Given matrix is \[A = \left[ {\begin{array}{*{20}{c}}3&{ - 3}&4\\2&{ - 3}&4\\0&{ - 1}&1\end{array}} \right]\].
The cofactors of A are
\[{A_{11}} = \left| {\begin{array}{*{20}{c}}{ - 3}&4\\{ - 1}&1\end{array}} \right|\]
\[ = - 3 \cdot 1 - 4\left( { - 1} \right)\]
\[ = 1\]
\[{A_{12}} = - \left| {\begin{array}{*{20}{c}}2&4\\0&1\end{array}} \right|\]
\[ = - \left( {2 \cdot 1 - 4 \cdot 0} \right)\]
\[ = - 2\]
\[{A_{13}} = \left| {\begin{array}{*{20}{c}}2&3\\0&{ - 1}\end{array}} \right|\]
\[ = 2 \cdot \left( { - 1} \right) - 3 \cdot 0\]
\[ = - 2\]
\[{A_{21}} = - \left| {\begin{array}{*{20}{c}}{ - 3}&4\\{ - 1}&1\end{array}} \right|\]
\[ = - \left[ { - 3 \cdot 1 - 4 \cdot \left( { - 1} \right)} \right]\]
\[ = - 1\]
\[{A_{22}} = \left| {\begin{array}{*{20}{c}}3&4\\0&1\end{array}} \right|\]
\[ = \left[ {3 \cdot 1 - 4 \cdot 0} \right]\]
\[ = 3\]
\[{A_{23}} = - \left| {\begin{array}{*{20}{c}}3&{ - 3}\\0&{ - 1}\end{array}} \right|\]
\[ = - \left[ {3 \cdot \left( { - 1} \right) - \left( { - 3} \right) \cdot 0} \right]\]
\[ = 3\]
\[{A_{31}} = \left| {\begin{array}{*{20}{c}}{ - 3}&4\\{ - 3}&4\end{array}} \right|\]
\[ = \left[ { - 3 \cdot 4 - \left( { - 3} \right) \cdot 4} \right]\]
\[ = 0\]
\[{A_{32}} = - \left| {\begin{array}{*{20}{c}}3&4\\2&4\end{array}} \right|\]
\[ = - \left[ {3 \cdot 4 - 2 \cdot 4} \right]\]
\[ = - 4\]
\[{A_{33}} = \left| {\begin{array}{*{20}{c}}3&{ - 3}\\2&{ - 3}\end{array}} \right|\]
\[ = \left[ {3 \cdot \left( { - 3} \right) - 2 \cdot \left( { - 3} \right)} \right]\]
\[ = - 3\]
The Adjoint of the matrix is
\[Adj\,A = \left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{21}}}&{{A_{31}}}\\{{A_{12}}}&{{A_{22}}}&{{A_{32}}}\\{{A_{13}}}&{{A_{23}}}&{{A_{33}}}\end{array}} \right]\]
Substitute the value of cofactors
\[ = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&0\\{ - 2}&3&{ - 4}\\{ - 2}&3&{ - 3}\end{array}} \right]\]
Hence option B is the correct option.
Additional information:
We can find the adjoint of a matrix if the matrix is a square matrix. This means the number of rows is equal to the number of columns. The adjoint of a matrix is used to the determine inverse of that matrix. In other words, we can find the inverse of a matrix if the matrix is a square matrix.
Note: Students often do a common mistake to find the adjoint of the matrix. They forgot to transpose the cofactor matrix. To calculate the adjoint we have to find the transpose matrix of the cofactor matrix.
Recently Updated Pages
JEE Advanced 2021 Physics Question Paper 2 with Solutions

Solutions Class 12 Notes JEE Advanced Chemistry [PDF]

JEE Advanced 2022 Chemistry Question Paper 2 with Solutions

JEE Advanced 2025 Revision Notes for Chemistry Energetics - Free PDF Download

JEE Advanced Marks vs Rank 2025 - Predict IIT Rank Based on Score

JEE Advanced 2022 Maths Question Paper 2 with Solutions

Trending doubts
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

JEE Advanced 2025 Notes

IIT CSE Cutoff: Category‐Wise Opening and Closing Ranks

IIT Fees Structure 2025

Top IIT Colleges in India 2025

Other Pages
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Instantaneous Velocity - Formula based Examples for JEE
